A series for Euler's constant $\gamma$
Messenger of Mathematics, XLVI, 1917, 73 – 80

1. In a paper recently published in this Journal (Vol.XLIV, pp. 1 – 10), Dr. Glaisher proves a number of formulĂ¦ of the type $$\gamma = 1 - 2 \left(\frac{S_3}{3\cdot 4} + \frac{S_5}{5\cdot 6} + \frac{S_7}{7\cdot 8} + \cdots \right),$$ where $$S_n = 1^{-n} + 2^{-n} + 3^{-n} + 4^{-n} + \cdots,$$ and conjectures the existence of a general formula $$\gamma = \lambda_r - (r+1) (r+2) \cdots (2r)$$ $$\times \left\{\frac{S_3}{3(r+3)(r+4) \cdots(2r+2)} + \frac{S_5}{5(r+5)(r+6)\cdots(2r+4)} + \cdots \right\},$$ where $\lambda_r$ is a rational number. I propose now to prove the general formula of which Dr. Glaisher's are particular cases: this formula is itself a particular case of still more general formulĂ¦.

2. Let $r$ and $t$ be any two positive numbers. Then

\begin{eqnarray} && \int\limits^1_0 x^{r-1} (1-x)^{t-1} \log \Gamma(1-x) \ dx = \int\limits^1_0 x^{t-1}(1-x)^{r-1} \log \Gamma (x) \ dx \nonumber \\ && =\int\limits^1_0 x^{t-1} (1-x)^{r-1} \log \Gamma(1+x) \ dx - \int\limits^1_0 x^{t-1} (1-x)^{r-1} \log x \ dx \end{eqnarray}
But
\begin{eqnarray} && \int\limits^1_0 x^{r-1} (1-x)^{t-1} \log \Gamma (1-x) \ dx \nonumber \\ &&= \int\limits^1_0 x^{r-1} (1-x)^{t-1} \left\{\gamma x + S_2\frac{x^2}{2}+S_3 \frac{x^3}{3} + \cdots \right\} \ dx \nonumber \\ &&=\frac{\Gamma(1+r)\Gamma(t)}{\Gamma(1+r+t)} \gamma + \frac{\Gamma(2+r) \Gamma(t)}{\Gamma(2+r+t)} \frac{S_2}{2}+\frac{\Gamma(3+r)\Gamma(t)}{\Gamma(3+r+t)}\frac{S_3}{3}+\cdots \end{eqnarray}
Similarly
\begin{eqnarray*} && \int\limits^1_0 x^{t-1} (1-x)^{r-1} \log \Gamma(1+x) \ dx \\ && = -\frac{\Gamma(1+t)\Gamma(r)}{\Gamma(1+r+t)} \gamma + \frac{\Gamma(2+t) \Gamma(r)}{\Gamma(2+r+t)} \frac{S_2}{2} -\frac{\Gamma(3+t)\Gamma(r)}{\Gamma(3+r+t)} \frac{S_3}{3} + \cdots \tag{2'} \end{eqnarray*}
And also
\begin{eqnarray} \int\limits^1_0 x^{t-1} (1-x)^{r-1} \log x \ dx &=& \frac{d}{dt} \int\limits^1_0 x^{t-1} (1-x)^{r-1} \ dx = \frac{d}{dt} \left\{\frac{\Gamma(t)\Gamma(r)}{\Gamma(r+t)}\right\}\nonumber \\ &=& \frac{\Gamma(r)\Gamma(t)}{\Gamma(r+t)} \left\{\frac{\Gamma'(t)}{\Gamma(t)}-\frac{\Gamma'(r+t)}{\Gamma(r+t)}\right\} \nonumber \\ &=& -\frac{\Gamma(r)\Gamma(t)}{\Gamma(r+t)} \int\limits^1_0 x^{t-1} \frac{1-x^r}{1-x} \ dx \end{eqnarray}
It follows from (1)(3) that, if $r$ and $t$ are positive, then
\begin{eqnarray} && \frac{r}{1(r+t)}\gamma+ \frac{r(r+1)}{2(r+t)(r+t+1)} S_2+ \frac{r(r+1)(r+2)}{3(r+t)(r+t+1)(r+t+2)} S_3 + \cdots \nonumber \\ &&+ \frac{t}{1(r+t)}\gamma - \frac{t(t+1)}{2(r+t)(r+t+1)}S_2 + \frac{t(t+1)(t+2)}{3(r+t)(r+t+1)(r+t+2)} S_3 - \cdots \nonumber \\ &&= \int\limits^1_0 \frac{x^{t-1}(1-x^r)}{1-x} \ dx \end{eqnarray}
Now, interchanging $r$ and $t$ in (4), and taking the sum and the difference of the two results, we see that, if $r$ and $t$ are positive, then
\begin{eqnarray} && \frac{r+t}{1(r+t)} \gamma + \frac{r(r+1)(r+2)+t(t+1)(t+2)}{3(r+t)(r+t+1)(r+t+2)}S_3 + \cdots \nonumber \\ &&= \frac{1}{2} \int\limits^1_0 \frac{x^{r-1} + x^{t-1} - 2x^{r+t-1}}{1-x} \ dx; \end{eqnarray}
and
\begin{eqnarray} \frac{r(r+1)-t(t+1)}{2(r+t)(r+t+1)}S_2 +\frac{r(r+1)(r+2)(r+3)-t(t+1)(t+2)(t+3)}{4(r+t)(r+t+1)(r+t+2)(r+t+3)} S_4 &&+ \cdots \nonumber \\ && = \frac{1}{2}\int\limits^1_0 \frac{x^{t-1} - x^{r-1}}{1-x} \ dx. \end{eqnarray}
The right-hand sides of (5) and (6) can be expressed in finite terms if $r$ and $t$ are rational. If, in particular, $r$ and $t$ are integers, then \begin{eqnarray*} \int\limits^1_0 \frac{x^{r-1}+x^{t-1} - 2x^{r+t-1}}{1-x} \ dx &&= \frac{1}{r} + \frac{1}{r+1} + \frac{1}{r+2} + \cdots +\frac{1}{r+t-1} \\ &&+ \frac{1}{t} + \frac{1}{t+1} + \frac{1}{t+2} + \cdots + \frac{1}{r+t-1}; \end{eqnarray*} and \begin{eqnarray*} \int\limits^1_0 \frac{x^{t-1}-x^{r-1}}{1-x} \ dx &&= \left(1+\frac{1}{2} + \frac{1}{3} + \frac{1}{4} + \cdots + \frac{1}{r-1}\right)\\ &&- \left(1 + \frac{1}{2} + \frac{1}{3}+\frac{1}{4}+ \cdots + \frac{1}{t-1}\right). \end{eqnarray*}

3. Let us now suppose that $t=r$ in (5). Then it is clear that

\begin{eqnarray} \gamma + \frac{(r+1)(r+2)}{3(2r+1)(2r+2)} S_3 &&+ \frac{(r+1)(r+2)(r+3)(r+4)}{5(2r+1)(2r+2)(2r+3)(2r+4)} S_5 + \cdots\nonumber \\ &&= \int\limits^1_0 \frac{x^{r-1}(1-x^r)}{1-x} \ dx = \int\limits^1_0 \frac{1+x^{2r-1}}{1+x} \ dx, \end{eqnarray}
if $r > 0$. If we suppose, in (7), that $r$ is an integer, we obtain the formula conjectured by Dr Glaisher, the value of $\lambda_r$ being $$\int\limits^1_0 \frac{1+x^{2r-1}}{1+x} \ dx = 1 - \frac{1}{2} + \frac{1}{3} - \frac{1}{4} + \cdots + \frac{1}{2r-1}.$$ Again, dividing both sides in (6) by $r-t$ and making $t \to r$, we see that, if $r > 0$, then
\begin{eqnarray} && \frac{r+1}{2(2r+1)} \left(\frac{1}{r} + \frac{1}{r+1}\right) S_2 \nonumber\\ &&+ \frac{(r+1)(r+2)(r+3)}{4(2r+1)(2r+2)(2r+3)} \left(\frac{1}{r} + \frac{1}{r+1}+ \frac{1}{r+2} + \frac{1}{r+3}\right) S_4 + \cdots \nonumber \\ &&\qquad= - \int\limits^1_0 \frac{x^{r-1} \log x}{1-x} \ dx = \frac{1}{r^2} +\frac{1}{(r+1)^2} + \frac{1}{(r+2)^2} + \frac{1}{(r+3)^2} + \end{eqnarray}
Thus for example we have $$\frac{\pi^2}{12} = (1 + \myfrac{1}{2}) \frac{S_2}{2\cdots 3} + (1+\myfrac{1}{2}+ \myfrac{1}{3}+\myfrac{1}{4}) \frac{S_4}{4\cdot 5} (1+\myfrac{1}{2} + \myfrac{1}{3}+ \myfrac{1}{4}+\myfrac{1}{5}+\myfrac{1}{6}) \frac{S_6}{6\cdot 7}+ \cdots.$$

4. If we start with the integral $$\int\limits^1_0 x^{r-1} (1-x)^{t-1} \log \Gamma \left(1 - \frac{x}{2}\right) \ dx,$$ and proceed as in $\S$ 2, we can shew that, if $r$ and $t$ are positive, then

\begin{eqnarray} \frac{r}{1(r+t)}&& S'_1 + \frac{r(r+1)}{2(r+t)(r+t+1)} S'_2 + \frac{r(r+1)(r+2)}{3(r+t)(t+t+1)(r+t+2)} S'_3 + \cdots \nonumber \\ &&- \frac{t}{1(r+t)} S'_1 + \frac{t(t+1)}{2(r+t)(r+t+1)} S'_2 - \frac{t(t+1)(t+2)}{3(r+t)(r+t+1)(r+t+2)} S'_3 + \cdots \nonumber \\ &&= \int\limits^1_0 \frac{x^{t-1}(1-x^r)}{1-x} \ dx - \log \frac{\pi}{2}, \end{eqnarray}
where $$S'_n = 1^{-n} - 2^{-n} + 3^{-n} - 4^{-n} + \cdots$$ From (9) we can easily deduce that, if $r$ and $t$ are positive, then
\begin{eqnarray} && \frac{r(r+1)+t(t+1)}{2(r+t)(r+t+1)} S'_2 \nonumber \\ &&+ \frac{r(r+1)(r+2)(r+3)+t(t+1)(t+2)(t+3)}{4(r+t)(r+t+1)(r+t+2)(r+t+3)} S'_4 + \cdots\nonumber \\ &&= \frac{1}{2} \int\limits^1_0 \frac{x^{r-1} + x^{t-1} - 2x^{r+t-1}}{1-x} \ dx - \log \frac{\pi}{2}; \end{eqnarray}
and
$$\frac{r-t}{1(r+t)} S'_1 + \frac{r(r+1)(r+2)-t(t+1)(t+2)}{3(r+t)(r+t+1)(r+t+2)} S'_3 + \cdots = \frac{1}{2} \int\limits^1_0 \frac{x^{t-1}- x^{r-1}}{1-x} \ dx .$$
As particular cases of (10) and (11), we have
$$\log \frac{\pi}{2} + \frac{r+1}{2(2r+1)} S'_2 + \frac{(r+1)(r+2)(r+3)}{4(2r+1)(2r+2)(2r+3)} S'_4 + \cdots = \int\limits^1_0 \frac{1+x^{2r-1}}{1+x} \ dx;$$
and
\begin{eqnarray} \frac{1}{r} S'_1 &&+ \frac{(r+1)(r+2)}{3(2r+1)(2r+2)} \left(\frac{1}{r} + \frac{1}{r+1} + \frac{1}{r+2} \right) S'_3 \nonumber \\ &&+ \frac{(r+1)(r+2)(r+3)(r+4)}{5(2r+1)(2r+2)(2r+3)(2r+4)} \nonumber \\ &&\times\left(\frac{1}{r}+ \frac{1}{r+1} + \frac{1}{r+2} + \frac{1}{r+3} + \frac{1}{r+4} \right) S'_5 + \cdots \nonumber \\ &&= \frac{1}{r^2} + \frac{1}{(r+1)^2} + \frac{1}{(r+2)^2} + \cdots, \end{eqnarray}
provided that $r \gt 0$. Thus for example we have \begin{eqnarray*} && 1 = \log \frac{\pi}{2} + 2 \left(\frac{S'_2}{2.3} + \frac{S'_4}{4.5} + \frac{S'_6}{6.7} + \cdots \right); \\ \frac{\pi^2}{12} = \frac{S'_1}{1.2} + \frac{S'_3}{3.4} &&(1+ \frac{1}{2} + \frac{1}{3}) + \frac{S'_5}{5.6} (1+\frac{1}{2} + \frac{1}{3} + \frac{1}{4} + \frac{1}{5}) + \cdots. \end{eqnarray*}

5. The preceding results may be generalised as follows. Let $\zeta(s,x)$ denote the function represented by the series $$x^{-s} + (x+1)^{-s} + (x+2)^{-s} + (x+3)^{-s} + \cdots (x \gt 0)$$ and its analytical continuations, so that $\zeta(s,1) = \zeta(s)$ and $\zeta(s, \frac{1}{2}) = (2^s - 1) \zeta(s), \zeta(s)$ being the Riemann $\zeta$-function. Then

\begin{eqnarray} \int\limits^1_0 && x^{r-1} (1-x)^{t-1} \zeta(s,1-x) \ dx = \int\limits^1_0 x^{t-1} (1-x)^{r-1} \zeta(s,x) \ dx \nonumber \\ &&= \int\limits^1_0 x^{t-1} (1-x)^{r-1} \zeta(s,1 + x) \ dx + \int\limits^1_0 x^{t-s-1}(1-x)^{r-1} \ dx, \end{eqnarray}
provided that $r$ and $t$ are positive. But we know that, if $|x|\lt 1$, then
$$\zeta(s,1 -x) = \zeta(s) + \frac{s}{1!} \zeta(s+1) x + \frac{s(s+1)}{2!} \zeta(s+2) x^2 + \cdots;$$
and that
$$\int\limits^1_0 x^{t-s-1} (1-x)^{r-1} \ dx = \frac{\Gamma(t-s)\Gamma(r)}{\Gamma(r-s+t)},$$
provided that $t > s$. It follows from (14)(16) that, if $r$ and $t$ are positive and $t > s$, then
\begin{eqnarray} && \left\{\zeta(s) + \frac{s}{1!} \frac{r}{r+t} \zeta(s+1) + \frac{s(s+1)}{2!} \frac{r(r+1)}{(r+t)(r+t+1)} \zeta(s+2) + \cdots\right\}\nonumber \\ &&- \left\{\zeta(s) - \frac{s}{1!}\frac{t}{r+t} \zeta(s+1) + \frac{s(s+1)}{2!} \frac{t(t+1)}{(r+t)(r+t+1)} \zeta(s+2) - \cdots \right\}\nonumber \\ &&= \frac{\Gamma(r+t) \Gamma(t-s)}{\Gamma(t) \Gamma(r-s+t)}. \end{eqnarray}
As particular cases of (17), we have
\begin{eqnarray} \frac{s}{1!}\frac{r+t}{r+t} \zeta(s+1) + \frac{s(s+1)(s+2)}{3!} && \frac{r(r+1)(r+2)+t(t+1)(t+2)}{(r+t)(r+t+1)(r+t+2)} \zeta(s+3) + \cdots \nonumber \\ &&= \frac{1}{2} \frac{\Gamma(r+t)}{\Gamma(r-s+t)} \left\{\frac{\Gamma(t-s)}{\Gamma(t)} + \frac{\Gamma(r-s)}{\Gamma(r)} \right\}, \end{eqnarray}
and
\begin{eqnarray} \frac{s(s+1)}{2!} && \frac{r(r+1)-t(t+1)}{(r+t)(r+t+1)} \zeta(s+2)\nonumber \\ &&+\frac{s(s+1)(s+2)(s+3)}{4!} \frac{r(r+1)(r+2)(r+3)-t(t+1)(t+2)(t+3)}{(r+t)(r+t+1)(r+t+2)(r+t+3)} \zeta(s+4)+ \cdots \nonumber \\ &&= \frac{1}{2} \frac{\Gamma(r+t)}{\Gamma(r-s+t)} \left\{\frac{\Gamma(t-s)}{\Gamma(t)} - \frac{\Gamma(r-s)}{\Gamma(r)}\right\}, \end{eqnarray}
provided that $r$ and $t$ are positive and greater than $s$. From (18) and (19) we deduce that, if $r$ is positive and greater than $s$, then
\begin{eqnarray} \frac{s}{1!} \frac{r}{2r} \zeta(s+1) + \frac{s(s+1)(s+2)}{3!} \frac{r(r+1)(r+2)}{2r(2r+1)(2r+2)} && \zeta(s+3) + \cdots\nonumber \\ = \frac{1}{2} \frac{\Gamma(2r)\Gamma(r-s)}{\Gamma(r)\Gamma(2r-s)}, \end{eqnarray}
and
\begin{eqnarray} && \frac{s(s+1)}{2!} \frac{r(r+1)}{2r(2r+1)} \left(\frac{1}{r} + \frac{1}{r+1} \right)\zeta(s+2)\nonumber \\ && +\frac{s(s+1)(s+2)(s+3)}{4!} \frac{r(r+1)(r+2)(r+3)}{2r(2r+1)(2r+2)(2r+3)}\nonumber \\ && \times \left(\frac{1}{r} + \frac{1}{r+1} + \frac{1}{r+2} + \frac{1}{r+3}\right) \zeta(s+4) + \cdots \nonumber \\ && = \frac{1}{2} \frac{\Gamma(2r)\Gamma(r-s)}{\Gamma(r)\Gamma(2r-s)} \int\limits^1_0 \frac{x^{r-s-1}(1-x^s)}{1-x} \ dx. \end{eqnarray}

6. If we start with the integral $$\int\limits^1_0 x^{r-1} (1-x)^{t-1} \zeta \left(s,1 - \frac{x}{2}\right) \ dx,$$ and proceed as in $\S$ 5, we can shew that, if $r$ and $t$ are positive and $t > s$, then

\begin{eqnarray} \zeta_1 (s) && + \frac{s}{1!} \frac{r}{r+t} \zeta_1(s+1) \frac{s(s+1)}{2!} \frac{r(r+1)}{(r+t)(r+t+1)}\zeta_1(s+2)+ \cdots\nonumber \\ && + \zeta_1 (s) - \frac{s}{1!} \frac{t}{r+t} \zeta_1(s+1) + \frac{s(s+1)}{2!} \frac{t(t+1)}{(r+t)(r+t+1)} \zeta_1 (s+2) - \cdots \nonumber \\ && = \frac{\Gamma(r+t)\Gamma(t-s)}{\Gamma(t)\Gamma(r-s+t)}, \end{eqnarray}
where $\zeta_1(s)$ is the function represented by the series $$1^{-s} - 2^{-s} + 3^{-s} - 4^{-s} + \cdots$$ and its analytical continuations. From (22) we deduce that, if $r$ and $t$ are positive and greater than $s$, then
\begin{eqnarray} (1+1) \zeta_1(s) &&+ \frac{s(s+1)}{2!} \frac{r(r+1)+t(t+1)}{(r+t)(r+t+1)} \zeta_1(s+2) + \cdots \nonumber \\ &&= \frac{1}{2} \frac{\Gamma(r+t)}{\Gamma(r-s+t)} \left\{\frac{\Gamma(t-s)}{\Gamma(t)} + \frac{\Gamma(r-s)}{\Gamma(r)}\right\}; \end{eqnarray}
and
\begin{eqnarray} &&\frac{s}{1!} \frac{r-t}{r+t} \zeta_1 (s+1)\nonumber \\ &+& \frac{s(s+1)(s+2)}{3!} \frac{r(r+1)(r+2)-t(t+1)(t+2)}{(r+t)(r+t+1)(r+t+2)} \zeta_1(s+3) + \cdots \nonumber \\ &=& \frac{1}{2} \frac{\Gamma(r+t)}{\Gamma(r-s+t)} \left\{\frac{\Gamma(t-s)}{\Gamma(t)} - \frac{\Gamma(r-s)}{\Gamma(r)}\right\}. \end{eqnarray}
As particular cases of (23) and (24), we have
$$\zeta_1(s) + \frac{s(s+1)}{2!} \frac{r(r+1)}{2r(2r+1)} \zeta_1(s+2) + \cdots = \frac{1}{2} \frac{\Gamma(2r)\Gamma(r-s)}{\Gamma(r) \Gamma(2r-s)},$$
and
\begin{eqnarray} \frac{s}{1!} \frac{r}{2r} \frac{1}{r} \zeta_1(s+1) &+& \frac{s(s+1)(s+2)}{3!} \frac{r(r+1)(r+2)}{2r(2r+1)(2r+2)} \left(\frac{1}{r} + \frac{1}{r+1} + \frac{1}{r+2} \right) \zeta_1 (s+3) + \cdots\nonumber \\ &=&\frac{1}{2} \frac{\Gamma(2r)\Gamma(r-s)}{\Gamma(r)\Gamma(2r-s)} \int\limits^1_0 \frac{s^{r-s-1}(1-x^s)}{1-x} \ dx, \end{eqnarray}
provided that $r$ is positive and greater than $s$.